In a mathematics class, half of the students scored 92 on an achievement test. With the exception of a few students who scored 54, the remaining students scored 73. Which statement about the distribution is true?

Step-by-step solution

1. 1

   Introduce counts. Let $N$ be the (even) class size: $\tfrac{N}{2}$ students scored $92$, $k \ge 1$ students scored $54$, and $\tfrac{N}{2} - k$ scored $73$. Treating "a few" as an unknown $k$ keeps the argument valid for every reading of the word.

2. 2

   Find the median. In ascending order the first $k$ values are $54$, the next $\tfrac{N}{2} - k$ are $73$, and the top $\tfrac{N}{2}$ are $92$. Position $\tfrac{N}{2}$ therefore holds $73$ and position $\tfrac{N}{2}+1$ holds $92$, so

   $\text{median} = \frac{73 + 92}{2} = 82.5$

   again independent of $k$.

3. 3

   Compute the mean.

   $\text{mean} = \frac{92\cdot\tfrac{N}{2} + 54k + 73\left(\tfrac{N}{2} - k\right)}{N} = \frac{82.5N - 19k}{N} = 82.5 - \frac{19k}{N}$

4. 4

   Compare the two. With $k \ge 1$ the correction $-\tfrac{19k}{N}$ is strictly negative:

   $\text{mean} = 82.5 - \frac{19k}{N} < 82.5 = \text{median}$

   The distribution is skewed left — a small tail well below the bulk pulls the mean down without moving the median. Option C is correct.

5. 5

   Check the other options. B is false since the mean is strictly below $82.5$. D reverses the inequality. A is false too: the mode is $92$ (half the class), and the mean is at most $82.5$, so the mean is less than the mode, not greater.

6. 6

   Verify with numbers. $N = 20$, $k = 3$: ten $92$s, three $54$s, seven $73$s give mean $\tfrac{920 + 162 + 511}{20} = 79.65$ against median $82.5$. $N = 100$, $k = 7$: mean $81.17$, median $82.5$. Both agree with $82.5 - \tfrac{19k}{N}$.